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2 years ago

A pizza shop sells pizzas that are 10 inches (in diameter) or larger. A 10-inch cheese pizza costs

Mathematics

2 answers:

Bogdan [553]2 years ago

5 0

Answer:

A pizza shop sells pizzas that are 10 inches orr larger. A 10-inch cheese pizza costs $8 and each additional costs $1.50 and each additional topping costs $0.75.

The equation that represents the cost of a pizza is:

P = $10 + $1.50a + $0.75b

Where 'a' represents the number additional inches and 'b' represents the number of additional toppings.

Lina20 [59]2 years ago

4 0

Answer: $T=8+ 1.50 x+0.75y$ , where 'x' denotes the number of additional inch to the pizza and 'y' denotes the number of additional toppings.

Step-by-step explanation:

Let 'x' denotes the number of additional inch to the pizza and 'y' denotes the number of additional toppings.

Given : A 10-inch cheese pizza costs $8.

i.e. Fixed cost = $8

Each additional inch costs $1.50, and each additional topping costs $0.75.

∴ Cost of additional inch = $1.50 x

∴ Cost of additional topping = $0.75y

Then The Total cost of any pizza would be :

Total cost = Fixed cost + Cost of additional inch + cost of additional topping

= $8+ $1.50 x+$0.75y

⇒ Equation that represents the cost of a pizza :

$T=8+ 1.50 x+0.75y$

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During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

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The first number is 824. This digits are stable numbers because 2 as a number is situated in the same place as the number (2nd position).

Hence, there are only 1 stable 3-digit numbers in the value 78247 since only a value exists as 2 in the value and there is no 1 and 3 in the value.

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Answer:

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Answer:

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